設(shè)點P(x,y),則√[x^2+(y-3)^2]-|y|=3,所以√[x^2+(y-3)^2]=|y|+3,x^2+y^2-6y+9=y^2+6|y|+9
所以x^2=6|y|+6y
當y>=0時,x^2=12y;
當y<0時,x^2=0,即x=0
所以軌跡方程為分段函數(shù)x^2=12y(y>=0)；x=0(y<0).